Question

1. Are the mean number of times a month a person eats out the same for whites, blacks, Hispanics and Asians?

White	Black	Hispanic	Asian
6	4	7	8
8	1	3	3
2	5	5	5
4	2	4	1
6		6	7

1. d_f numerator and denominator =
2. Write H_o and H_a
3. What is the p-value?
4. Do you reject or accept the null hypothesis at the 5% level?

Answer 1

The following table is obtained:

	White	Black	Hispanic	Asian
	6	4	7	8
	8	1	3	3
	2	5	5	5
	4	2	4	1
	6		6	7
Sum =	26	12	25	24
Average =	5.2	3	5	4.8
	156	46	135	148
St. Dev. =	2.28	1.826	1.581	2.864
SS =	20.8	10	10	32.8
n =	5	4	5	5

The total sample size is N = 19. Therefore, the total degrees of freedom are:

df_{total} = 19 - 1 = 18

Also, the between-groups degrees of freedom are df_{between} = 4 - 1 = 3 , and the within-groups degrees of freedom are:

df_{within} = df_{total} - df_{between} = 18 - 3 = 15

First, we need to compute the total sum of values and the grand mean. The following is obtained

Also, the sum of squared values is

Based on the above calculations, the total sum of squares is computed as follows

The within sum of squares is computed as shown in the calculation below:

The between sum of squares is computed directly as shown in the calculation below:

Now that sum of squares are computed, we can proceed with computing the mean sum of squares:

Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows:

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ₁ = μ₂ = μ₃ = μ₄

Ha: Not all means are equal

The above hypotheses will be tested using an F-ratio for a One-Way ANOVA.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df_1 = 3 and df_2 = 3, therefore, the rejection region for this F-test is

(3) Test Statistics

(4) Decision about the null hypothesis

Since it is observed that F = 0.885 ≤Fc​=3.287, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.4711 , and since p = 0.4711 ≥0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that not all 4 population means are equal, at the α=0.05 significance level.